Answer:
Step-by-step explanation:
1. 8 days
2. C?
3. A
4. B
5. C
6.D
7.48
8. A
9. 1.69%
10. False
I have no clue if these are right i tried my best to get them right, if they are wrong I am incredibly sorry!
- SavageSavvy
That is when h=0
so assuming yo meant
h(t)=-16t²+24t+16
solve for t such that h(t)=0
because when height=0, the gymnast hits the ground
so
0=-16t²+24t+16
using math (imma complete the square
0=-16(t²-3/2t)+16
0=-16(t²-3/2t+9/16-9/16)+16
0=-16((t-3/4)²-9/16)+16
0=-16(t-3/4)²+9+16
0=-16(t-3/4)²+25
-25=-16(t-3/4)²
25/16=(t-3/4)²
sqrt both sides
+/-5/4=t-3/4
3/4+/-5/4=t
if we do plus (because minus would give us negative height)
8/4=t
2=t
it takes 2 seconds
Answer:
45 min.
Step-by-step explanation:
1 1/2 hours = 90 min.
90 min. / 2 brothers = 45 min. each
Answer:
D) a chi square test for independence.
Step-by-step explanation:
Given that we suspect that automobile insurance premiums (in dollars) may be steadily decreasing with the driver's driving experience (in years), so we choose a random sample of drivers who have similar automobile insurance coverage and collect data about their ages and insurance premiums.
We are to check whether two variables insurance premiums and driving experience are associated.
Two categorical variables are compared for different ages and insurance premiums.
Hence a proper test would be
D) a chi square test for independence.
Answer:
Step-by-step explanation:
Hello!
X: number of absences per tutorial per student over the past 5 years(percentage)
X≈N(μ;σ²)
You have to construct a 90% to estimate the population mean of the percentage of absences per tutorial of the students over the past 5 years.
The formula for the CI is:
X[bar] ± 
* 
⇒ The population standard deviation is unknown and since the distribution is approximate, I'll use the estimation of the standard deviation in place of the population parameter.
Number of Absences 13.9 16.4 12.3 13.2 8.4 4.4 10.3 8.8 4.8 10.9 15.9 9.7 4.5 11.5 5.7 10.8 9.7 8.2 10.3 12.2 10.6 16.2 15.2 1.7 11.7 11.9 10.0 12.4
X[bar]= 10.41
S= 3.71
[10.41±1.645*
]
[9.26; 11.56]
Using a confidence level of 90% you'd expect that the interval [9.26; 11.56]% contains the value of the population mean of the percentage of absences per tutorial of the students over the past 5 years.
I hope this helps!
